Tìm x trong các hỗn số sau:
a) 2\(\frac{x}{7}\)= \(\frac{75}{35}\)
b) 4\(\frac{3}{x}\)= \(\frac{47}{x}\)
c) x\(\frac{x}{15}\)=\(\frac{112}{5}\)
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a) 2\(\frac{x}{7}\) = \(\frac{75}{35}\)
\(\frac{2.7+x}{7}\) = \(\frac{75:5}{35:5}\) = \(\frac{15}{7}\)
=> 2.7+x = 15
14+x = 15
x = 15-14 = 1
Vậy x=1
b)4\(\frac{3}{x}\) = \(\frac{47}{x}\)
\(\frac{4.x+3}{x}\) = \(\frac{47}{x}\)
=> 4.x + 3 = 47
4x= 47-3=44
vậy x= 44:4=11
c)x\(\frac{x}{15}\) = \(\frac{112}{5}\)
x\(\frac{x}{15}\) =\(\frac{112.3}{5.3}\) = \(\frac{336}{15}\)
\(\frac{x.15+x.1}{15}\) = \(\frac{336}{15}\)
=>(15+1) x =336
16x = 336
x = 336 : 16
vậy x = 21
\(a,2\frac{x}{7}=\frac{75}{35}\)
\(\frac{75}{35}=2\frac{5}{35}=2\frac{1}{7}\)
\(\Rightarrow x=1\)
\(b,47:4=...\left(+3\right)\)
\(\Rightarrow\left(47-3\right):4=11\)
\(\Rightarrow x=11\)
\(c,\frac{112}{5}=\frac{336}{15}=22\left(+6\right)\)
\(\Rightarrow x=22\frac{x=6}{15}\)
\(d,7,5x=\frac{75x}{100}:\left(9-6\frac{13}{21}\right)=2\frac{13}{25}\)
\(\frac{75x}{100}:\left(9-\frac{139}{21}\right)=\frac{63}{25}\)
\(\frac{75x}{100}:\frac{50}{21}=\frac{63}{25}\)
\(\Rightarrow\frac{63^3}{25^{\left(1\right)}}\cdot\frac{50^{\left(2\right)}}{21^1}=6=\frac{6}{1}\)
\(\Rightarrow\frac{6}{1}=\frac{600}{100}\)
Con d sai đề
\(a,4\frac{3}{x}=\frac{47}{x}\)
\(\frac{4x+3}{x}=\frac{47}{x}\)
\(\Rightarrow4x^2+3x=47x\)
\(4x^2=47x-3x=44x\)
\(4x^2:4x=44x:4x\)
\(x=11\)
\(b,x\frac{x}{15}=\frac{112}{5}\)
\(\frac{15x+x}{15}=\frac{112}{5}\)
\(75x+5x=112.15\)
\(80x=1680\)
\(x=21\)
a, \(4\frac{3}{x}=\frac{47}{x}\)
\(\Leftrightarrow\frac{4x+3}{x}=\frac{47}{x}\)
\(\Leftrightarrow4x+3=47\)
\(4x=47-3\)
\(4x=44\)
\(x=44:4\)
\(x=11\)
b, \(x\frac{x}{15}=\frac{112}{5}\)
\(\Leftrightarrow\frac{15x+x}{15}=\frac{112}{5}\)
\(\Leftrightarrow\frac{16x}{15}=\frac{336}{15}\)
\(\Leftrightarrow16x=336\)
\(x=336:16\)
\(x=21\)
a, \(\frac{1}{6}x+\frac{1}{10}-\frac{4}{15}x+1=0\)
\(\Leftrightarrow-\frac{1}{10}x=-\frac{11}{10}\)
\(\Leftrightarrow x=11\)
b,\(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\Leftrightarrow\frac{1}{7}x-\frac{2}{7}=0\)hoặc \(-\frac{1}{5}x+\frac{3}{5}=0\)hoặc \(\frac{1}{3}x+\frac{4}{3}=0\)
+) \(\frac{1}{7}x-\frac{2}{7}=0\Leftrightarrow\frac{1}{7}x=\frac{2}{7}\Leftrightarrow x=2\)
+)\(-\frac{1}{5}x+\frac{3}{5}=0\Leftrightarrow-\frac{1}{5}x=-\frac{3}{5}\Leftrightarrow x=3\)
+)\(\frac{1}{3}x+\frac{4}{3}=0\Leftrightarrow\frac{1}{3}x=-\frac{4}{3}\Leftrightarrow x=-4\)
c, \(\frac{1}{2}x-\frac{11}{15}:\frac{33}{35}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{9}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}x=\frac{4}{9}\)
\(\Leftrightarrow x=\frac{8}{9}\)
a/ \(\frac{1}{6}x+\frac{1}{10}-\frac{4}{15}x+1=0\)
\(\Rightarrow-\frac{1}{10}x=-\frac{11}{10}\)
\(\Rightarrow x=11\)
b/ \(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\Rightarrow\frac{1}{7}x-\frac{2}{7}=0\Rightarrow\frac{1}{7}x=\frac{2}{7}\Rightarrow x=2\)
hoặc \(-\frac{1}{5}x+\frac{3}{5}=0\Rightarrow-\frac{1}{5}x=-\frac{3}{5}\Rightarrow x=3\)
hoặc \(\frac{1}{3}x+\frac{4}{3}=0\Rightarrow\frac{1}{3}x=-\frac{4}{3}\Rightarrow x=-4\)
Vậy x = 2, x = 3, x = -4
c/ \(\frac{1}{2}x-\frac{11}{15}:\frac{33}{35}=-\frac{1}{3}\)
\(\Rightarrow\frac{1}{2}x-\frac{7}{9}=-\frac{1}{3}\)
\(\Rightarrow\frac{1}{2}x=\frac{4}{9}\Rightarrow x=\frac{8}{9}\)
Vậy x = 8/9
a) Ta có:
\(\frac{4}{15}+\frac{1}{6}-\frac{4}{9}>\frac{2}{3}-x-\frac{1}{4}\\ \Rightarrow x+\frac{4}{15}+\frac{1}{6}-\frac{4}{9}>\frac{2}{3}-\frac{1}{4}\\ \Rightarrow x>\frac{2}{3}+\frac{4}{9}-\frac{1}{4}-\frac{1}{6}-\frac{4}{15}\\ \Rightarrow x>\left(\frac{6}{9}+\frac{4}{9}\right)-\left(\frac{15}{60}+\frac{10}{60}+\frac{16}{60}\right)\)
\(x>\frac{10}{9}-\frac{41}{60}\\ x>\frac{200-123}{180}\Rightarrow x>\frac{77}{180}\)
b) Bất đẳng thức kép
\(4-1\frac{1}{3}< x+\frac{1}{5}< 12\frac{2}{7}-3\frac{3}{8}\)
có nghĩa là ta phải có hai bất đẳng thức đồng thời:
\(x+\frac{1}{5}>4-1\frac{1}{3}\) và \(x+\frac{1}{5}< 12\frac{2}{7}-3\frac{3}{8}\)
Ta tìm các giá trị của x cần thỏa mãn bất đẳng thức thứ nhất:
\(x+\frac{1}{5}>4-1\frac{1}{3}\Rightarrow x>4-1\frac{1}{3}-\frac{1}{5}\\ \Rightarrow x>\frac{37}{15}\)
Từ bất đẳng thức thứ hai
\(x+\frac{1}{5}< 12\frac{2}{7}-3\frac{3}{8}\Rightarrow x< \frac{86}{7}-\frac{27}{8}-\frac{1}{5}\\ \Rightarrow x< \frac{2439}{280}.\)
Như vậy các số hữu tỉ x cần thỏa mãn:
\(\frac{37}{15}< x< \frac{2439}{280}\)
a) \(4\frac{3}{x}=\frac{47}{x}\)
\(\Rightarrow\frac{4\times x+3}{x}=\frac{47}{x}\)
\(\Rightarrow4\times x+3=47\)
\(4\times x=47-3\)
\(4\times x=44\)
\(x=44:4=11\)
b) \(x\frac{x}{15}=\frac{112}{5}\)
\(\Rightarrow\frac{x\times15+x}{15}=\frac{336}{15}\)
\(\Rightarrow\frac{x\times16}{15}=\frac{336}{15}\)
\(\Rightarrow x\times16=336\)
\(x=21\)
Chúc bn học tốt!!!!
\(\frac{1+0,6-\frac{3}{7}}{\frac{8}{3}+\frac{8}{5}-\frac{8}{7}}=\frac{\frac{3}{3}+\frac{3}{5}-\frac{3}{7}}{\frac{8}{3}+\frac{8}{5}-\frac{8}{7}}=\frac{3.\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\right)}{8.\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\right)}=\frac{3.1}{8.1}=\frac{3}{8}\)
\(\frac{\frac{1}{3}+0,25-\frac{1}{5}+0,125}{\frac{7}{6}+\frac{7}{8}-0,7+\frac{7}{16}}=\frac{\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}}{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}+\frac{7}{16}}=\frac{1.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}\right)}{7.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}\right)}=\frac{1.1}{7.1}=\frac{1}{7}\)
=>\(\frac{3}{8}-\frac{1}{7}=\frac{13}{56}\)
\(2\frac{x}{7}=\frac{75}{35}\)
\(\frac{2.7+x}{7}=\frac{75}{35}\)
\(\frac{14+x}{7}=\frac{75}{35}\)
\(\frac{5\left(14+x\right)}{35}=\frac{75}{35}\)
=>5 ( 14 + x ) = 75
14 + x = 75 : 5
14 + x = 15
x = 15 - 14
x = 1
Vậy x =1